All categories

2 years ago

Hypothesis: How will molecular size affect the rate of evaporation and how will this correlate

1 answer:

5 0

A molecular size affects the rate of evaporation when the larger the intermolecular forces in a compound, the slower the evaporation rate and this correlates with temperature change.

Molecular size seems to have an effect on evaporation rates in that the larger a molecule gets or grows from a base chemical formula, its evaporation rate will get slower.

<h3>What is the molecular size?</h3>

This is a measure of the area a molecule occupies in three-dimensional space as this relates to the physical size of an individual molecule.

Hence, we can see that a molecular size affects the rate of evaporation the larger the forces, the lower the rate.

Read more about<em> molecular size</em> here:

brainly.com/question/16616599

#SPJ1

You might be interested in

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of $N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol$

amount of $SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol$

mole fraction of $N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330$

mole of fraction $SF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669$

PV = nRT

P of N_2 O $= \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa$

mole of SF_4 $=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa$

Total pressure = 39127.053 + 79126.36 = 118253.413 Pa

6 0

3 years ago

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinatio

Phantasy [73]

I guess B is the right one......

4 0

3 years ago

Read 2 more answers

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago

The reaction below is at equilibrium. What would happen if more sodium<br> were added?

densk [106]

Answer:

d is your answer please follow me

7 0

1 year ago

Read 2 more answers

Salt is an ionic compount. it

ElenaW [278]

Answer:

Explanation:

Ionic Compounds Are Balanced

Table salt is an example of an ionic compound. Sodium and chlorine ions come together to form sodium chloride, or NaCl. The sodium atom in this compound loses an electron to become Na+, while the chlorine atom gains an electron to become Cl-

3 0

2 years ago